Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $x = \dfrac{8y^2 + 8y}{9y - 54} \times \dfrac{-3y + 18}{y^2 + 7y + 6} $
Solution: First factor the quadratic. $x = \dfrac{8y^2 + 8y}{9y - 54} \times \dfrac{-3y + 18}{(y + 1)(y + 6)} $ Then factor out any other terms. $x = \dfrac{8y(y + 1)}{9(y - 6)} \times \dfrac{-3(y - 6)}{(y + 1)(y + 6)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ 8y(y + 1) \times -3(y - 6) } { 9(y - 6) \times (y + 1)(y + 6) } $ $x = \dfrac{ -24y(y + 1)(y - 6)}{ 9(y - 6)(y + 1)(y + 6)} $ Notice that $(y - 6)$ and $(y + 1)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ -24y\cancel{(y + 1)}(y - 6)}{ 9(y - 6)\cancel{(y + 1)}(y + 6)} $ We are dividing by $y + 1$ , so $y + 1 \neq 0$ Therefore, $y \neq -1$ $x = \dfrac{ -24y\cancel{(y + 1)}\cancel{(y - 6)}}{ 9\cancel{(y - 6)}\cancel{(y + 1)}(y + 6)} $ We are dividing by $y - 6$ , so $y - 6 \neq 0$ Therefore, $y \neq 6$ $x = \dfrac{-24y}{9(y + 6)} $ $x = \dfrac{-8y}{3(y + 6)} ; \space y \neq -1 ; \space y \neq 6 $